2.1 INSTANTANEOUS RATE OF CHANGE
1
2.1 Instantaneous Rate of Change
In this section, we discuss the concept of the instantaneous rate of change of
a given function. As an application, we use the velocity of a moving object.
The motion of an object along a line at a particular instant is very difficult
to define precisely. The modern approach consists of computing the average
velocity over smaller and smaller time intervals. To be more precise, let s(t)
be the position function or displacement of a moving object at time t. We
would like to compute the velocity of the object at the instant t = t0 .
Average Velocity
We start by finding the average velocity of the object over the time interval
t0 ≤ t ≤ t0 + ∆t given by the expression
v=
Distance T raveled
s(t0 + ∆t) − s(t0 )
=
.
Elapsed T ime
∆t
(2.1.1)
Geometrically, the average velocity over the time interval [t0 , t0 + ∆t] is just
the slope of the line joining the points (t0 , s(t0 )) and (t0 + ∆t, s(t0 + ∆t)) on
the graph of s(t).(See Figure 2.1.1)
Figure 2.1.1
Example 2.1.1
A freely falling body experiencing no air resistance falls s(t) = 16t2 feet in t
seconds. Complete the following table
time interval
[1.8,2] [1.9,2] [1.99,2] [1.999,2] [2,2.0001] [2,2.001] [2,2.01]
Average velocity
2
Solution.
From time t = 1.8 to time t = 2, formula (2.1.1) gives
s(2) − s(1.8)
−16(2)2 − [−16(1.8)2 ]
=
= −60.8 ft/sec.
2 − 1.8
0.2
Using formula (2.1.1) on each of the remaining intervals, we find
time interval
[1.8,2] [1.9,2] [1.99,2] [1.999,2] [2,2.0001] [2,2.001] [2,2.01]
Average velocity 60.8
62.4
63.84
63.98
64.0016
64.016
64.16
Instantaneous Velocity and Speed
The next step is to calculate the average velocity on smaller and smaller time
intervals ( that is, make ∆t close to zero). The average velocity in this case
approaches what we would intuitively call the instantaneous velocity at
time t = t0 which is defined using the limit notation by
s(t0 + ∆t) − s(t0 )
.
∆t→0
∆t
v(t0 ) = lim
(2.1.2)
Geometrically, the instantaneous velocity at t0 is the slope of the tangent
line to the graph of s(t) at the point (t0 , s(t0 )).(See Figure 2.1.2)
Figure 2.1.2
Example 2.1.2
For the distance function in Example 2.1.1, find the instantaneous velocity
at t = 2.
2.1 INSTANTANEOUS RATE OF CHANGE
3
Solution.
Examining the bottom row of the table in Example 2.1.1, we see that the average velocity seems to be approaching the value 64 as we shrink the time intervals. Thus, it is reasonable to expect the velocity to be v(2) = 64 f t/sec
We define the speed of a moving object to be the absolute value of the
velocity function. Sometimes there is confusion between the words “speed”
and “velocity”. Speed is a nonnegative number that indicates how fast an
object is moving, whereas velocity indicates both speed and direction(relative
to a coordinate system). For example, if the object is moving along a vertical line we define a positive velocity when the object is going upward and a
negative velocity when the object is going downward.
If the position function s(t) is replaced by an arbitrary function f (x) then we
define the instantaneous rate of change of a function y = f (x) at x = a
to be
f (x) − f (a)
(2.1.3)
lim
x→a
x−a
where
f (x) − f (a)
x−a
(2.1.4)
is the average rate of change of f (x) from x = a to x = b, also known as
the difference quotient of f (x) from x = a to x = b.
The instantaneous rate of change of a function f (x) from x = a to x = a + h
is called the derivative of f at a and we denote it by f 0 (a) :
f (a + h) − f (a)
.
h→0
h
f 0 (a) = lim
(2.1.5)
If the derivative exists, i.e., can be found, then we say that the function is
differentiable at x = a. The process of finding the derivative of a function
is called differentiation. If a function has no derivative at a point then we
say that it is non-differentiable there.
Since the instantaneous rate of change represents the slope of a tangent line,
f 0 (a) is the slope of the tangent line to the graph of y = f (x) at the point
(a, f (a)). The equation of the tangent line is given by
y − f (a) = f 0 (a)(x − a).
(2.1.6)
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Example 2.1.3
(a) Find f 0 (1) for f (x) = x2 .
(b) Find the equation of the tangent line to the graph of f (x) at the point
(1, f (1)).
Solution.
Completing the following chart
x
f (b)−f (a)
b−a
[0.9,1] [0.99,1] [0.999,1] [1,1.0001] [1,1.001] [1,1.01] [1,1.1]
1.9
1.99
1.999
2.0001
2.001
2.01
2.1
we see that f 0 (1) = 2.
(b) The equation of the tangent line is
y − f (1) = f 0 (1)(x − 1)
or
y − 1 = 2(x − 1).
In point-intercept form, we have y = 2x − 1
Example 2.1.4 (Numerical Estimation of the Derivative)
Find approximate values for f 0 (x) at each of the x−values given in the following table
x
0
5
f (x) 100 70
10 15 20
55 46 40
Solution.
The derivative can be estimated by using the average rate of change or the
difference quotient
f (a + h) − f (a)
.
f 0 (a) ≈
h
If a is a left-endpoint then f 0 (a) is estimated by
f 0 (a) ≈
f (b) − f (a)
b−a
where b > a. If a is a right-endpoint then f 0 (a) is estimated by
f 0 (a) ≈
f (a) − f (b)
a−b
2.1 INSTANTANEOUS RATE OF CHANGE
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where b < a. If a is an interior point then f 0 (a) is estimated by
1 f (a) − f (b) f (c) − f (a)
0
f (a) ≈
+
2
a−b
c−a
where b < a < c. For example,
f 0 (0) ≈
f 0 (5) ≈
f 0 (10) ≈
f 0 (15) ≈
f 0 (20) ≈
f (5) − f (0)
= −6
5
1 f (10) − f (5) f (5) − f (0)
+
= −4.5
2
5
5
1 f (15) − f (10) f (10) − f (5)
+
= −2.4
2
5
5
1 f (20) − f (15) f (15) − f (10)
+
= −1.5
2
5
5
f (20) − f (15)
= −1.2.
5
(5)
The quantity f (10)−f
is known as the right slope estimation of f 0 (5).
10−5
Similarly, we can estimate f 0 (5) by using a left slope estimation,i.e.,
f 0 (5) ≈
f (5) − f (0)
= −6.
5−0
An improved estimation consists of taking the average of the left slope and
the right slope, that is,
f 0 (5) ≈
−3 − 6
= −4.5
2